Consider a long thin conducting wire carrying a uniform current I. A particle having mass "M" and charge "q" is released at a distance "a" from the wire with a speed $$\epsilon_{\circ}$$ along the direction of current in the wire. The particle gets attracted to the wire due to magnetic force. The particle turns round when it is at distance x from the wire. The value of x is [$$\mu_{\circ}$$ is vacuum permeability]

Let the wire lie along the z-axis with current $$I \hat{k}$$.

The particle is in the x-z plane at distance $$r$$ from the wire.

$$B = \frac{\mu_0 I}{2\pi r} (-\hat{j})$$

Taking coordinates $$(r, z)$$ where $$r$$ is the perpendicular distance from the wire:

$$\vec{v} = v_r \hat{i} + v_z \hat{k}$$

$$\vec{F} = q(\vec{v} \times \vec{B}) = q \left[(v_r \hat{i} + v_z \hat{k}) \times \left(-\frac{\mu_0 I}{2\pi r} \hat{j}\right)\right] = -\frac{q\mu_0 I v_r}{2\pi r}\hat{k} + \frac{q\mu_0 I v_z}{2\pi r}\hat{i}$$

Equation of motion along the z-direction:

$$M \frac{dv_z}{dt} = -\frac{q\mu_0 I}{2\pi r} v_r \implies M \frac{dv_z}{dt} = -\frac{q\mu_0 I}{2\pi r} \frac{dr}{dt}$$

$$M dv_z = -\frac{q\mu_0 I}{2\pi} \frac{dr}{r}$$

Integrating from the initial state ($$r=a, v_z=v_0$$) to the turning point ($$r=x$$ where $$v_r=0$$):

Since magnetic force does no work, total speed remains $$v_0$$. At the turning point, $$v_r = 0 \implies v_z = -v_0$$.

$$\int_{v_0}^{-v_0} M dv_z = -\frac{q\mu_0 I}{2\pi} \int_{a}^{x} \frac{dr}{r}$$

$$-2M v_0 = -\frac{q\mu_0 I}{2\pi} \ln\left(\frac{x}{a}\right) \implies \ln\left(\frac{x}{a}\right) = -\frac{4\pi M v_0}{q\mu_0 I}$$

$$x = a e^{-\frac{4\pi M v_0}{q\mu_0 I}}$$